Solve the equation. $\dfrac{dy}{dx}=\dfrac{(xy)^2}{8}+y^2$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=-\dfrac{x^2+16}{16+C}$ (Choice B) B $y=-\dfrac{16}{x^2+16+C}$ (Choice C) C $y=-\dfrac{x^3+24x}{24+C}$ (Choice D) D $y=-\dfrac{24}{x^3+24x+C}$
Solution: We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{(xy)^2}{8}+y^2 \\\\ \dfrac{dy}{dx}&=\dfrac{y^2}{8}(x^2+8) \\\\ \dfrac{8}{y^2}\,dy&=(x^2+8)\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} \dfrac{8}{y^2}\,dy&=(x^2+8)\,dx \\\\ \int \dfrac{8}{y^2}\,dy&=\int (x^2+8)\,dx \\\\ -\dfrac{8}{y}&=\dfrac{x^3}{3}+8x+C_1 \\\\ \dfrac{1}{y}&=-\dfrac{x^3}{24}-x+C_2 \\\\ \dfrac{1}{y}&=-\dfrac{x^3+24x+C}{24} \\\\ y&=-\dfrac{24}{x^3+24x+C} \end{aligned}$ [Where did we get C?] Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=-\dfrac{24}{x^3+24x+C}$